Correct Answers
x equals 6 and x equals negative 3
$$ x=6 \quad \text{and} \quad x=-3 $$
Explanation
From the given equation:fraction with numerator of x and denominator of x plus 6 equals fraction with numerator of 2 x minus 3 and denominator of x plus 12.
$$ \frac{x}{x+6} = \frac{2x-3}{x+12} $$
to first eliminate the fractions, we can cross-multiply and expand both sides.
x open parenthesis x plus 12 close parenthesis equals open parenthesis 2 x minus 3 close parenthesis open parenthesis x plus 6. $$ x(x + 12) = (2x - 3)(x + 6) $$
x squared plus 12 x equals 2 x squared plus 12 x minus 3 x minus 18. $$ x^2 + 12x = 2x^2 + 12x - 3x - 18 $$
x squared plus 12 x equals 2 x squared plus 9 x minus 18. $$ x^2 + 12x = 2x^2 + 9x - 18 $$
Use inverse operations to get an expression equal to zero, simplify, and distribute the negative sign. After combining like terms, you may want to multiply both sides by negative 1 \( -1 \) to make the equation a little easier to work with.x squared plus 12 x minus open parenthesis 2 x squared plus 9 x minus 18 close parenthesis equals 0.
$$ x^2 + 12x - (2x^2 + 9x - 18) = 0 $$
x squared plus 12 x minus 2 x squared minus 9 x plus 18 equals 0. $$ x^2 + 12x - 2x^2 - 9x + 18 = 0 $$
the opposite of x squared plus 3 x plus 18 equals 0. $$ -x^2 + 3x + 18 = 0 $$
x squared minus 3 x minus 18 equals 0. $$ x^2 - 3x - 18 = 0 $$
We can then factor the quadratic equation. We need two numbers that multiply to negative 18 \( -18 \) and add up to negative 3 \( -3. \) These numbers are negative 6 \( -6 \) and 3:x squared minus 3 x minus 18 equals open parenthesis x minus 6 close parenthesis open parenthesis x plus 3 close parenthesis, equals 0.
$$ x^2 - 3x - 18 = (x - 6)(x + 3) = 0 $$
The solutions, based on these factors x minus 6 equals 0 or x plus 3 equals 0: \( (x - 6 = 0 \text{ or } x + 3 = 0): \) x equals 6 or x equals negative 3.
$$ x = 6 \quad \text{or} \quad x = -3 $$
One last thing: Since the original problem involved fractions, we need to make sure none of the solutions are extraneous. Extraneous solutions would make a denominator in the original equation 0, because division by zero is undefined. If x equals 6, \( x = 6, \) the denominators x plus 6 equals 12 \( x+6 = 12 \) and x plus 12 equals 18 \( x+12 = 18 \) are both nonzero, so x equals 6 \( x=6 \) is valid. If x equals negative 3, \( x = -3, \) the denominators x plus 6 equals 3 \( x+6 = 3 \) and x plus 12 equals 9 \( x+12 = 9 \) are both nonzero, so x equals negative 3 \( x=-3 \) is valid as well.